krit.club logo

Stoichiometry - Reacting masses and volumes

Grade 11IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The mole is the unit of amount of substance. One mole contains Avogadro's constant (6.02imes10236.02 imes 10^{23}) of particles.

•

The Relative Molecular Mass (MrM_r) is the sum of the Relative Atomic Masses (ArA_r) of the elements in a formula.

•

Molar mass is the mass of one mole of a substance and is expressed in g/molg/mol. It is numerically equal to the MrM_r.

•

Stoichiometry involves using the coefficients of a balanced chemical equation to determine the mole ratio of reactants and products.

•

At Room Temperature and Pressure (r.t.p.), one mole of any gas occupies a volume of 24extdm324 ext{ dm}^3 (which is 24,000extcm324,000 ext{ cm}^3).

•

Concentration of a solution is defined as the number of moles of solute per unit volume of solution (mol/dm3mol/dm^3).

•

The Limiting Reactant is the reactant that is completely consumed in a reaction, determining the maximum amount of product formed.

•

Percentage yield is calculated by Actual YieldTheoretical Yield×100%\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%, representing the efficiency of a reaction.

📐Formulae

n=mMrn = \frac{m}{M_r}

n=V24 (where V is in dm3 at r.t.p.)n = \frac{V}{24} \text{ (where } V \text{ is in } dm^3 \text{ at r.t.p.)}

n=c×V (where V is in dm3)n = c \times V \text{ (where } V \text{ is in } dm^3 \text{)}

Percentage Yield=Actual YieldTheoretical Yield×100%\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

Percentage Purity=Mass of pure substanceTotal mass of sample×100%\text{Percentage Purity} = \frac{\text{Mass of pure substance}}{\text{Total mass of sample}} \times 100\%

💡Examples

Problem 1:

Calculate the mass of CO2CO_2 produced when 10extg10 ext{ g} of calcium carbonate (CaCO3CaCO_3) is heated until it completely decomposes according to the equation: CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g). (Ar:Ca=40,C=12,O=16A_r: Ca=40, C=12, O=16)

Solution:

Mr of CaCO3=40+12+(3×16)=100M_r \text{ of } CaCO_3 = 40 + 12 + (3 \times 16) = 100. Moles of CaCO3=10extg100extg/mol=0.1extmolCaCO_3 = \frac{10 ext{ g}}{100 ext{ g/mol}} = 0.1 ext{ mol}. From the equation, the ratio of CaCO3:CO2CaCO_3 : CO_2 is 1:11:1. Therefore, moles of CO2=0.1extmolCO_2 = 0.1 ext{ mol}. Mr of CO2=12+(2×16)=44M_r \text{ of } CO_2 = 12 + (2 \times 16) = 44. Mass of CO2=0.1extmol×44extg/mol=4.4extgCO_2 = 0.1 ext{ mol} \times 44 ext{ g/mol} = 4.4 ext{ g}.

Explanation:

First, find the moles of the known substance using its MrM_r. Use the balanced equation ratio to find the moles of the unknown substance, then convert those moles back to mass.

Problem 2:

What volume of hydrogen gas, H2H_2, measured at r.t.p., is required to react completely with 12extdm312 ext{ dm}^3 of oxygen gas, O2O_2, to form water? 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

Solution:

According to Avogadro’s Law, the volume ratio of gases is the same as the mole ratio at constant temperature and pressure. The ratio of H2:O2H_2 : O_2 is 2:12:1. Volume of H2=2×Volume of O2H_2 = 2 \times \text{Volume of } O_2. Volume of H2=2×12extdm3=24extdm3H_2 = 2 \times 12 ext{ dm}^3 = 24 ext{ dm}^3.

Explanation:

For gas-only calculations at the same temperature and pressure, you can use the stoichiometric coefficients directly as volume ratios without converting to moles first.

Problem 3:

25extcm325 ext{ cm}^3 of 0.2extmol/dm30.2 ext{ mol/dm}^3 NaOHNaOH reacts with HClHCl. Calculate the number of moles of NaOHNaOH used.

Solution:

Volume in dm3=251000=0.025extdm3dm^3 = \frac{25}{1000} = 0.025 ext{ dm}^3. Moles n=c×V=0.2extmol/dm3×0.025extdm3=0.005extmoln = c \times V = 0.2 ext{ mol/dm}^3 \times 0.025 ext{ dm}^3 = 0.005 ext{ mol}.

Explanation:

When dealing with solutions, always ensure the volume is converted from cm3cm^3 to dm3dm^3 by dividing by 10001000 before using the concentration formula.

Reacting masses and volumes - Revision Notes & Key Formulas | IGCSE Grade 11 Chemistry